Algebraic delta functions Let $D_n$ be the Weyl algebra: $D_n \subset \mathrm{End}_\mathbf{C}(\mathbf{C}[x_1, \dots, x_n])$ is generated by $x_i$ and $\partial_i$, $i = 1, \dots, n$.  My professor says the $D_n$-module $$ M = \frac{\mathbf{C}[x_1, \dots, x_n][x_1^{-1}]}{\mathbf{C}[x_1, \dots, x_n]} $$ describes an algebraic version of the delta function.  In my notes I have written 'delta functions and their derivatives'.   Specifically, he said the $D_n$-module $$M_1 = \frac{\mathbf{C}[x][x^{-1}]}{\mathbf{C}[x]} $$ can be thought of as a delta function sitting at the point $x=0$, and that more generally the module $M$ can be thought of as a family of delta functions sitting at each point of the hyperplane $x_1 = 0$.  
I understand that as a complex vector space $M_1$ is generated by $\frac{1}{x}, \frac{1}{x^2}, \dots$  And the $D_n$-action is fully described by $$x \cdot \frac{1}{x} = 0, \;\;\;\;\; x \cdot \frac{1}{x^n} = \frac{1}{x^{n-1}}, n > 1; \;\;\;\;\; \partial \cdot \frac{1}{x^n} = -  \frac{n}{x^{n+1}}, n \geq 1.  $$  And similarly for $M$, which is generated over $\mathbf{C}$ by the Laurent polynomials vanishing at infinity with coefficients in $\mathbf{C}[x_2, \dots, x_n]$.  
Can someone help me better understand what my professor means?  Thanks.  
 A: Let's stick to the case $n = 1$. The Dirac delta $\delta$ at the origin generates a Weyl algebra module, and you want to know what module it is. In this algebraic setting, the defining property of $\delta$ is that
$$\int p(x) \delta(x) \, dx = p(0)$$
where $p(x)$ is a polynomial. We can compute the derivatives of $\delta$ using repeated integration by parts: the first derivative is
$$\int p(x) \frac{\partial}{\partial x} \delta(x) \, dx = - \int \frac{\partial p}{\partial x} \delta(x) \, dx = -p'(0)$$
and in general the $n^{th}$ derivative sends $p$ to $(-1)^n \frac{\partial^n p}{\partial x^n}(0)$. We'll write this as $\delta^{(n)}(x)$. 
Now let's compute the action of $x$. By definition,
$$\int p(x) x \delta^{(n)}(x) \, dx = (-1)^n \int \frac{\partial^n}{\partial x^n} \left( x p(x) \right) \delta(x) \, dx.$$
Let's think for a minute about what this iterated derivative is. It consists of terms for each way to allocate derivatives to each of the factors $p(x)$ and $x$, by an iterated application of the product rule. But $x$ can only be differentiated once before it vanishes. So in fact there are only two terms, namely $x \frac{\partial^n p}{\partial x^n}$ and $n \frac{\partial^{n-1} p}{\partial x^{n-1}}$. The first term disappears after substituting $x = 0$, so only the second term is left, and we get
$$x \delta^{(n)}(x) = - n \delta^{(n-1)}(x)$$
and in particular $x \delta(x) = 0$. If we give this module the basis $\{ \delta(x), - \delta^{(1)}(x), \frac{1}{2} \delta^{(2)}(x), \dots, \frac{(-1)^n}{n!} \delta^{(n)}(x), \dots \}$ (the $n^{th}$ term here extracts the $n^{th}$ Taylor coefficient of a polynomial), then you can verify that sending $\frac{(-1)^n}{n!} \delta^{(n)}(x)$ to $\frac{1}{x^{n+1}}\in M_1$ is an isomorphism of Weyl algebra modules with what you wrote. (Except that your computation of the action of $\partial$ is incorrect.) 
